poj3292--Semi-prime H-numbers(数论篇2)

Semi-prime H-numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7680		Accepted: 3319

Description

This problem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory of 4n+1 numbers. Here, we do only a bit of that.

An H-number is a positive number which is one more than a multiple of four: 1, 5, 9, 13, 17, 21,... are the H-numbers. For this problem we pretend that these are the only numbers. The H-numbers are closed under multiplication.

As with regular integers, we partition the H-numbers into units, H-primes, and H-composites. 1 is the only unit. An H-number h is H-prime if it is not the unit, and is the product of two H-numbers in only one way: 1 × h. The rest of the numbers are H-composite.

For examples, the first few H-composites are: 5 × 5 = 25, 5 × 9 = 45, 5 × 13 = 65, 9 × 9 = 81, 5 × 17 = 85.

Your task is to count the number of H-semi-primes. An H-semi-prime is an H-number which is the product of exactly two H-primes. The two H-primes may be equal or different. In the example above, all five numbers are H-semi-primes. 125 = 5 × 5 × 5 is not an H-semi-prime, because it's the product of three H-primes.

Input

Each line of input contains an H-number ≤ 1,000,001. The last line of input contains 0 and this line should not be processed.

Output

For each inputted H-number h, print a line stating h and the number of H-semi-primes between 1 and h inclusive, separated by one space in the format shown in the sample.

Sample Input

21 
85
789
0

Sample Output

21 0
85 5
789 62

Source

Waterloo Local Contest, 2006.9.30

 

题目大意：给定1,4,9,13，，，，的数列叫做H-numbers，如果一个数仅能被两个H-H-semi-primes相乘（不能是1）得到，那么他就是H-semi-primes，问1到n内的H-semi-primes有多少个。

筛法求出来，vis初始为0.，代表它仅能有1*自身得到，为1代表是一个H-H-semi-primes数，为-1代表其他。

当两个i,j, vis[i] == 0 && vis[j] == 0 那么vis[i*j] = 1 ;

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
int vis[1100000] , a[1100000] , cnt ;
void init()
{
    int i , j ;
    memset(vis,0,sizeof(vis)) ;
    cnt = 0 ;
    for(i = 5 ; i <= 1010000 ; i += 4)
    {
        for(j = 5 ; j <= i ; j+= 4)
        {
            if( i*j > 1010000 ) break ;
            if( !vis[i] && !vis[j] ) vis[i*j] = 1 ;
            else
                vis[i*j] = -1 ;
        }
    }
    memset(a,0,sizeof(a)) ;
    for(i = 1 ; i < 1010000 ; i++)
    {
        a[i] = a[i-1] ;
        if( vis[i] == 1 )
            a[i]++ ;
    }
    return ;
}
int main()
{
    int n ;
    init() ;
    while( scanf("%d", &n) && n )
    {
        printf("%d %d\n", n, a[n]) ;
    }
    return 0;
}